Edit: The answer is no as Alex Bartel commented. So let me extend condition (b) to:
(b) $x_i \equiv b*x_j\pmod{p^m}$ for some integers $m,b$
(or even, $\alpha_1x_1 + ... + \alpha_nx_n \equiv 0\pmod{p^m}$)
Also, let me extend the question: If the answer is no, can the conditions be extended a bit for the answer to be yes? More generally, is there an easy way to describe all those subgroups?

@Lee: $(\mathbb{Z}/p\mathbb{Z})^n$ is a vector space over the field $\mathbb{Z}/p\mathbb{Z}$. You're looking for its subspaces. They can be described by equations "some coordinates are these linear combinations of the other coordinates". So you will need this type of conditions.
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user8268Apr 13 '11 at 11:35

To expand on the comment by user8268: sometimes, you will need more than one congruence. E.g., the set of $(x,y,z)$ in ${\bf Z}_p^3$ such that $x\equiv y\equiv z\pmod p$ is a subgroup that cannot be described by a single congruence.
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Gerry MyersonApr 13 '11 at 13:04

1

I don't understand the $p^{k_i}$ in front of the $\mathbb{Z}_{p^k}$. That just has the effect of making the cylic group a cyclic group of order $p^{k-k_i}$ instead of order $p^k$. Wouldn't it be much simpler to simply write that $G$ is a product of cyclic $p$-groups?
–
Arturo MagidinApr 13 '11 at 15:40

1 Answer
1

Yes, every subgroup can be characterized as the set of tuples satisfying certain linear congruences in its entries. To be more precise, we have the following:

Let $p$ be a prime. Then every subgroup of $\displaystyle \prod_{i=1}^n \mathbb{Z}_{p^k}$ (including the subgroups of a $G$ as given in the statement of the question) can be described as the set of elements $(h_1,\ldots,h_n)$ which satisfy a set of (at most) $n$ congruences of the following form:
$$\begin{align*}
h_{\sigma(1)} &\equiv 0 \pmod{p^{j_1}}\\
h_{\sigma(2)} &\equiv \alpha_{21}h_{\sigma(1)} \pmod{p^{j_2}}\\
h_{\sigma(3)} &\equiv \alpha_{31}h_{\sigma(1)} + \alpha_{32}h_{\sigma(2)} \pmod{p^{j_3}}\\
&\vdots\\
h_{\sigma(n)} &\equiv \alpha_{n1}h_{\sigma(1)} + \alpha_{n2}h_{\sigma(2)} + \cdots + \alpha_{n(n-1)}h_{\sigma(n-1)} \pmod{p^{j_n}}
\end{align*}$$
where $\sigma\in S_n$ is a permutation of $\{1,\ldots,n\}$, $0\leq j_1\leq\cdots\leq j_n\leq k$ are nonnegative integers, and $\alpha_{ij}$ are integers.

What follows is essentially the proof that submodules of free modules over PIDs are free, as described in this previous answer, particularized for this situation.

Let $H$ be a subgroup of $\displaystyle \prod_{i=1}^n \mathbb{Z}_{p^k}$.

We proceed by induction on $n$. If $n=1$, then $H$ is a subgroup of $\mathbb{Z}_{p^k}$, and therefore is of the form $p^{j_1}\mathbb{Z}_{p^k}$ for some $j_1$, $0\leq j_1 \leq k$; thus, $(h_1)\in H$ if and only if $h_1\equiv 0\pmod{p^{j_1}}$, which gives the result we want.

Asume the result holds for subgroups of a product of $n$ copies of $\mathbb{Z}_{p^k}$, and let $H$ be a subgroup of $\displaystyle \prod_{i=1}^{n+1}\mathbb{Z}_{p^k}$.

If $H=\{e\}$, then we are done: $H$ is described by the congruences $h_i\equiv 0\pmod{p^{k}}$, $i=1,\ldots,n$. So we may assume $H$ is nontrivial.

Consider the projections onto the $n+1$ coordinates; if any projection, say the projection on the $j$th component is trivial, then $H$ is a subgroup of a product of $n$ copies of $\mathbb{Z}_{p^k}$, and can be described as in the proposition, together with the condition $h_j\equiv 0\pmod{p^k}$.

Assume then that all projections are nontrivial. Each projection gives a subgroup of $\mathbb{Z}_{p^k}$, hence of the form $p^{r_i}\mathbb{Z}_{p^k}$, with $0\leq r_i\lt k$. Let $\sigma(1)$ be the index with the largest subgroup (i.e., smallest $r_i$, possibly $0$); set $j_1 = r_{\sigma(1)}$. Let $\mathbf{a}_1$ be an element of $H$ whose $\sigma(1)$ coordinate is $p^{j_1}$. The minimality of $j_1$ means that $\mathbf{a}_{1}$ can be written as
$$\mathbf{a}_1 = (a_1,\ldots,a_n) = p^{j_1}(b_1,\ldots,b_n) = a_{\sigma(1)}(b_1,\ldots,b_n)$$
with $b_{\sigma(1)}=1$, all $b_i$ integers.

Now let $K$ be the subgroup of $H$ corresponding to the kernel of the projection onto the $\sigma(1)$-coordinate; that is, all elements that have $\sigma(1)$ coordinate congruent to $0$ modulo $p^{k}$. By induction, there is a bijection $\sigma\colon \{2,\ldots,n+1\}\to\{1,\ldots,n+1\}-\{\sigma(1)\}$ and integers $\gamma_{rs}$, $2\leq r\leq n+1$, $1\leq s\lt r$ and nonnegative integers $j_2\leq\cdots\leq j_{n+1}\leq k$ such that the elements of $K$ can be described as
$$\begin{align*}
k_{\sigma(1)} &\equiv 0 \pmod{p^k}\\
k_{\sigma(2)} &\equiv 0 \pmod{p^{j_2}}\\
k_{\sigma(3)} &\equiv \gamma_{32}k_{\sigma(2)} \pmod{p^{j_3}}\\
&\vdots\\
k_{\sigma(n+1)} &\equiv \gamma_{(n+1)2}k_{\sigma(2)} + \cdots + \gamma_{(n+1)n}k_{\sigma(n)}\pmod{p^{j_{n+1}}}.
\end{align*}$$
Moreover, the minimality of $j_1$ ensures that $j_1\leq j_2$.

Now, $K\cap\langle \mathbf{a}_1\rangle = \{e\}$, and $K+\langle \mathbf{a}_1\rangle\subseteq H$. In fact, $H = K\oplus\langle\mathbf{a}_1\rangle$: given $\mathbf{h}\in H$, we know that the $\sigma(1)$-component of $\mathbf{h}$ is a multiple of $p^{j_1}$ (by choice of $\sigma(1)$), and so there exists $m$ such that $\mathbf{h}- m\mathbf{a}_1 \in K$. Hence, $h\in K\oplus\langle\mathbf{a}_1\rangle$.

Therefore, given $\mathbf{h}=(h_1,\ldots,h_{n+1})\in \displaystyle\prod_{i=1}^{n+1}\mathbb{Z}_{p^k}$, we have that $\mathbf{h}\in H$ if and only if:

However, $a_{\sigma(i)} = p^{j_1}b_{\sigma(i)} = a_{\sigma(1)}b_{\sigma(i)}$, and since $ma_{\sigma(1)} \equiv h_{\sigma(1)} \pmod{p^{k}}$ then for each $i$, $2\leq i\leq n+1$ we have
$$ma_{\sigma(i)} = ma_{\sigma(1)}b_{\sigma(i)} \equiv h_{\sigma(1)}b_{\sigma(i)}\pmod{p^{k}}.$$
Since $j_i\leq k$, we can then rewrite the congruences above as:
$$\begin{align*}
h_{\sigma(1)} &\equiv 0 \pmod{p^{j_1}}\\
h_{\sigma(2)} -h_{\sigma(1)}b_{\sigma(2)}&\equiv 0 \pmod{p^{j_2}}\\
h_{\sigma(3)} -h_{\sigma(1)}b_{\sigma(3)}
&\equiv \gamma_{32}\bigl(h_{\sigma(2)}-h_{\sigma(1)}b_{\sigma(2)}\bigr) \pmod{p^{j_3}}\\
&\vdots\\
h_{\sigma(n+1)}-h_{\sigma(1)}b_{\sigma(n+1)}
&\equiv \gamma_{(n+1)2}\bigl(h_{\sigma(2)}-h_{\sigma(1)}b_{\sigma(2)}\bigr) \\
&\qquad\qquad\mathop{+} \cdots
\mathop{+} \gamma_{(n+1)n}\bigl(h_{\sigma(n)}-h_{\sigma(1)}b_{\sigma(n)}\bigr)\pmod{p^{j_{n+1}}}
\end{align*}$$
which can be rewritten in the desired form by moving the $h_{\sigma(1)}b_{\sigma(k)}$ summand to the right hand side, and regrouping; this satisfies the conditions of the statement above, since $b_1,\ldots,b_n$ and $\gamma_{ij}$ are fixed, giving rise to fixed $\alpha_{ij}$ that do not depend on $\mathbf{h}$.

@Arturo, isn't that subgroup of ${\bf Z}^3$ given by the congruences $b\equiv6a\pmod9$, $c\equiv15a+2b\pmod{81}$?
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Gerry MyersonApr 14 '11 at 5:10

@Gerry: Hmmm... Yes. $2b = 9m+6a$, and plugging that for your $2b$ gives my $18m+27a$. So there may be a way, I'm just not seeing how to phrase it in general. Inductively, if I can figure out how to do it for the second coordinate, it should straightforward to go from there. I'll sleep on it. Thanks!
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Arturo MagidinApr 14 '11 at 5:20

If it's any help, here's how I got there: general element is $(1,6,27)r+(0,9,18)s+(0,0,81)t=(a,b,c)$ with $a=r$, $b=6r+9s=6a+9s$, $c=27r+18s+81t=15r+2(6r+9s)+81t=15a+2b+81t$, and the congruences pop out.
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Gerry MyersonApr 14 '11 at 6:55

@Gerry: Thanks. There's a few more simplifications that can be made (in the basis elements, you can always assume that any entry to the right of the "pivot" is a multiple of the pivot, for instance) which might make this doable in general.
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Arturo MagidinApr 14 '11 at 13:04

@Gerry: Thanks again; that was very useful, actually. I think I got it, if my "you may assume"s are not invalit somewhere around the line.
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Arturo MagidinApr 14 '11 at 21:40